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    <h1 class="post-title">
      Record_first_double_week_contest
    </h1>
  
  
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      <time datetime="2022-02-06T00:13:05&#43;08:00">
        
  February 6, 2022

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      <a class="category-link" href="https://zgjoget.gitee.io/categories/algorithm-training">algorithm training</a>, 
    
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              <p>记一次leetcode周赛和双周赛</p>
<h2 id="__25__-__2230-2359__-第71场双周赛httpsleetcode-cncomcontestbiweekly-contest-71-由理想汽车力扣联合主办"><strong>2.5</strong> <strong>22:30-23:59</strong> <a href="https://leetcode-cn.com/contest/biweekly-contest-71/">第71场双周赛</a> 由「理想汽车&amp;力扣」联合主办</h2>
<p>题目如下：</p>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_2.png" alt="pic2"></p>
<blockquote>
<p>这场是在正月初五的晚上，拜完年回家22:50才发现我报名了这一次竞赛，所以匆匆忙忙的打开电脑，实际答题的时间大约一小时。前两题都是比较简单不到半小时就解决了，第三题，我大概花了30分钟debug，后来在23:57左右提交发现超时了，后面时间就不够了。</p>
</blockquote>
<ul>
<li>
<ol start="2160">
<li><a href="https://leetcode-cn.com/problems/minimum-sum-of-four-digit-number-after-splitting-digits/">拆分数位后四位数字的最小和</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_3.png" alt="pic3"></p>
<pre tabindex="0"><code>class Solution {
public:
    int minimumSum(int num) {
        int new1, new2;
        vector&lt;int&gt;nums;
        for (int i = 0; i &lt; 4; i++) {
            nums.push_back(num % 10);
            num /= 10;
        }
        sort(nums.begin(), nums.end());
            int res, tmp;
            new1 = nums[0] * 10 + nums[3];
            new2 = nums[1] * 10 + nums[2];
            res = new1 + new2;
        return res;
    }
};
</code></pre><blockquote>
<p>这题比较简单，不过我在考虑时花了比较多的时间，其实在获得四位数之后，他们能组成的最小和就是ad+bc(设a&lt;b&lt;c&lt;d)</p>
</blockquote>
<ul>
<li>
<ol start="2161">
<li><a href="https://leetcode-cn.com/problems/partition-array-according-to-given-pivot/">根据给定数字划分数组</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_4.png" alt="pic4"></p>
<pre tabindex="0"><code>class Solution {
public:
    vector&lt;int&gt; pivotArray(vector&lt;int&gt;&amp; nums, int pivot) {
        vector&lt;int&gt;low,equal,large;
        for(int i=0;i&lt;nums.size();i++){
            if(nums[i]&gt;pivot)large.push_back(nums[i]);
            else if(nums[i]==pivot)equal.push_back(nums[i]);
            else low.push_back(nums[i]);
        }
        vector&lt;int&gt;res;
        for(int i=0;i&lt;low.size();i++){
            res.push_back(low[i]);
        }
        for(int i=0;i&lt;equal.size();i++){
            res.push_back(equal[i]);
        }
        for(int i=0;i&lt;large.size();i++){
            res.push_back(large[i]);
        }
        return res;
    }
};
</code></pre><blockquote>
<p>这题我是使用了low、equal、large，3个vector容器来存储遍历数组过程中出现的数据，最后根据顺序分别将三个容器中的元素弹出来。</p>
</blockquote>
<ul>
<li>
<ol start="2162">
<li><a href="https://leetcode-cn.com/problems/minimum-cost-to-set-cooking-time/">设置时间的最少代价</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_5.png" alt="pic5"></p>
<pre tabindex="0"><code>
</code></pre><blockquote>
<p>这题提交了，但是最后总是提示超时。</p>
</blockquote>
<ul>
<li>
<ol start="2163">
<li><a href="https://leetcode-cn.com/problems/minimum-difference-in-sums-after-removal-of-elements/">删除元素后和的最小差值</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_6.png" alt="pic6"></p>
<h2 id="__26__-1030-1159-第279场周赛httpsleetcode-cncomcontestweekly-contest-279新春虎年力扣官方招募专场"><strong>2.6</strong> 10:30-11:59 <a href="https://leetcode-cn.com/contest/weekly-contest-279/">第279场周赛</a>新春虎年「力扣官方招募」专场</h2>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_1.png" alt="pic1"></p>
<ul>
<li>
<ol start="2164">
<li><a href="https://leetcode-cn.com/problems/sort-even-and-odd-indices-independently/">对奇偶下标分别排序</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_7.png" alt="pic7"></p>
<pre tabindex="0"><code>class Solution {
public:
    vector&lt;int&gt; sortEvenOdd(vector&lt;int&gt;&amp; nums) {
        if(nums.size()&lt;=2)return nums;
        vector&lt;int&gt;a,b,res;
        for(int i=0;i&lt;nums.size();i=i+2){
            a.push_back(nums[i]);
        }
        for(int i=1;i&lt;nums.size();i=i+2){
            b.push_back(nums[i]);
        }
        sort(a.begin(),a.end());
        sort(b.begin(),b.end());
        int total=0;
        int p=0,q=b.size()-1;
        
        while(total&lt;nums.size()){
            res.push_back(a[p]);
            total++;
            if(total==nums.size())break;
            res.push_back(b[q]);
            total++;
            p++;
            q--;
        }
        return res;
    }
};
</code></pre><ul>
<li>
<ol start="6001">
<li><a href="https://leetcode-cn.com/problems/smallest-value-of-the-rearranged-number/">重排数字的最小值</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_8.png" alt="pic8"></p>
<pre tabindex="0"><code>class Solution {
public:
    long long smallestNumber(long long num) {
        int tag=0;
        if(num&lt;0){
            tag=1;
            num=-num;
        }
        else if(num==0)return 0;
        vector&lt;int&gt;st;
        while(num){
            st.push_back(num%10);
            num/=10;
        }
        sort(st.begin(),st.end());
        int start=0;
        while(st[start]==0)start++;
        long long res=0;
        if(tag==0){
            long long p=1;
            for(int i=st.size()-1;i&gt;=start+1;i--){
                res+=st[i]*p;
                p*=10;
            }
            for(int i=0;i&lt;start;i++)p*=10;
            res+=st[start]*p;
            return res;
        }
        else{
            long long q=pow(10,st.size()-1);
            for(int i=st.size()-1;i&gt;=start;i--){
                res+=st[i]*q;
                q/=10;
            }
            return -res;
        }
        }
};
</code></pre><ul>
<li>
<ol start="6002">
<li><a href="https://leetcode-cn.com/problems/design-bitset/">设计位集</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_9.png" alt="pic9">
最开始我直接就按照题目的意思做，最后也是要么超时、要么执行错误。代码如下：</p>
<pre tabindex="0"><code>class Bitset {
public:
    int *bit;
    int siz;
    Bitset(int size) {
        bit=new int[size];
        siz=size;
        memset(bit,0,size*4);
    }
    
    void fix(int idx) {
        bit[idx]=1;
    }
    
    void unfix(int idx) {
        bit[idx]=0;
    }
    
    void flip() {
        for(int i=0;i&lt;siz;i++){
            if(bit[i]==1)bit[i]=0;
            else bit[i]=1;
        }
    }
    
    bool all() {
        for(int i=0;i&lt;siz;i++){
            if(bit[i]!=1)return false;
        }
        return true;
    }
    
    bool one() {
        for(int i=0;i&lt;siz;i++){
            if(bit[i]==1)return true;
        }
        return false;
    }
    
    int count() {
        int cot=0;
        for(int i=0;i&lt;siz;i++){
            if(bit[i]==1)cot++;
        }
        return cot;
    }
    
    string toString() {
        string s;
        for(int i=0;i&lt;siz;i++)
        s+=to_string (bit[i]);
        return s;
    }
};

/**
 * Your Bitset object will be instantiated and called as such:
 * Bitset* obj = new Bitset(size);
 * obj-&gt;fix(idx);
 * obj-&gt;unfix(idx);
 * obj-&gt;flip();
 * bool param_4 = obj-&gt;all();
 * bool param_5 = obj-&gt;one();
 * int param_6 = obj-&gt;count();
 * string param_7 = obj-&gt;toString();
 */
</code></pre><p>考虑到，除了toString和flip这两个函数其他函数都是O(1)，flip()、toString()这两个函数是O(1)，而题目中限制了toString()函数的使用次数，所以我们应该考虑flip()函数的时间复杂度。
as we all know, 按位取反两次其实就变回去了，所以我们只要记录flip的执行次数，这里我是直接使用一个bool值来记录，最后提交通过！代码如下：</p>
<pre tabindex="0"><code>class Bitset {
    int n;
    int sum;
    bool flipped;
    vector&lt;bool&gt; v;
    
public:
    Bitset(int size) {
        n = size;
        sum = 0;
        flipped = false;
        v = vector&lt;bool&gt;(n);
    }
    
    void fix(int idx) {
        if (!(v[idx] ^ flipped))
            v[idx] = !v[idx], sum++;
    }
    
    void unfix(int idx) {
        if (v[idx] ^ flipped)
            v[idx] = !v[idx], sum--;
    }
    
    void flip() {
        flipped = !flipped;
        sum = n - sum;
    }
    
    bool all() {
        return sum == n;
    }
    
    bool one() {
        return sum &gt; 0;
    }
    
    int count() {
        return sum;
    }
    
    string toString() {
        string ans;
        for (int i = 0; i &lt; n; ++i)
            if (v[i] ^ flipped)
                ans.push_back('1');
            else
                ans.push_back('0');
        return ans;
    }
};

</code></pre><ul>
<li>
<ol start="2167">
<li><a href="https://leetcode-cn.com/contest/weekly-contest-279/problems/minimum-time-to-remove-all-cars-containing-illegal-goods/">移除所有载有违禁货物车厢所需的最少时间</a></li>
</ol>
</li>
</ul>
<p><img src="https://zgjoget9.github.io/img/double_week_contest_10.png" alt="pic10"></p>
<pre tabindex="0"><code>
</code></pre><h2 id="总结">总结</h2>
<p>这次双周会很能反应我的真实水平，遇到比较难一点的题，要么就完全做不出来，要么做出来时空要求不合格。另外前面简单题还要多加强，缩减消耗时间。</p>
              


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